Question

What volume of air contains 10.0g of oxygen gas at 273 k and 1.00 atm?

Answer 1

Answer: The volume of air will be 7.00 L

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$\text{Moles of oxygen gas}=\frac{10.0g}{32g/mol}=0.3125mol$

To calculate the volume of gas, we use the equation given by ideal gas that follows:

$PV=nRT$

where,

P = pressure of the gas  = 1.00 atm

V = Volume of the gas  = ?

T = Temperature of the gas = 273 K

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of gas = 003125 mol

Putting values in above equation, we get:

$1.00atm\times V=0.3125mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 273K\\\\V=\frac{0.3125\times 0.0821\times 273}{1.00}=7.00Ll$

Hence, the volume of air will be 7.00 L

Answer 2

12.3g 02times1 mol 02/32.0 g 02=mol 02
.38mol/x=20.95/100
x=1.8mol air... 

PV=nRT

101.3kPa times V =1.8 mol times 8.31[kPa *L]/{k*mol}*273k
V=40.3Lair